---
tag: 0HW7BNHHGF
kind: statement
role: theorem
name: Limit comparison test
msc: [40A05]
uses: [04Z5G0THN6, 0XKWSSY8P2]
sources:
  - url: https://en.wikipedia.org/wiki/Limit_comparison_test
    locator: Statement
same_as:
  wikidata: Q6549441
  wikipedia: Limit comparison test
---

## Statement

Let $(a_n)$ and $(b_n)$ be real sequences with $a_n \ge 0$ and $b_n > 0$ for all $n$.
Assume that $a_n / b_n \to c$ as $n \to \infty$, for a real number $c$ with $0 < c < \infty$.
Then $\sum a_n$ converges if and only if $\sum b_n$ converges.

## Remarks

For large $n$ we have $\frac{c}{2} b_n \le a_n \le 2c\, b_n$. The [[comparison test|04Z5G0THN6]] then gives both directions.
For example, $\frac{n}{n^3 + 1} / \frac{1}{n^2} \to 1$, so $\sum n/(n^3 + 1)$ converges, like $\sum 1/n^2$.
If $c = 0$, convergence of $\sum b_n$ still implies convergence of $\sum a_n$. The converse can fail.

## Explanations

### L1 human

If two series of positive terms have terms of the same size in the long run, they behave the same way.
So you can replace a complicated term by a simple one with the same leading behaviour.